Question

6. A certain quantity of helium gas is at a temperature of 27°C and a pressure of 1.00
atm. What will the new temperature be if its volume is doubled at the same time
that its pressure is decreased to one-half its original value? (Hint: Three variables
have been given so what equation will you use?)
I

Answer 1

Answer:

300K

Explanation:

Intitial temperature T1= 27°c +273= 300K

Initial pressure P1= 1.00 ATM

Initial volume V1= V

Final temperature T2= ???

Final volume V2 = 2V

Final pressure P2=P/2= 1/2= 0.5atm

How using the general gas equation, we have;

P1V1/T1=P2V2/T2

P1V1T2=P2V2T1

T2= P2V2T1/P1V1

SUBSTITUTING VALUES;

T2= 0.5 × 2V × 300/ 1 × V

T2= 300K